[proofplan] The proof is the parabolic [Bochner identity](/page/Bochner%20Identity) computation for the energy density of a [harmonic map heat flow](/page/Harmonic%20Map%20Heat%20Flow). The identity expresses $(\partial_t-\Delta_g)e(u)$ as the sum of a nonpositive second-derivative term, a domain [Ricci curvature](/page/Ricci%20Curvature) term, and a target [curvature tensor](/page/Riemann%20Curvature%20Tensor) term. Nonpositive sectional curvature of $N$ makes the target curvature contribution nonpositive, while compactness of $M$ gives a uniform lower bound for $\operatorname{Ric}_g$. Dropping the nonpositive terms and bounding the remaining Ricci contribution gives the required inequality. [/proofplan]

[step:Compute the parabolic Bochner identity for the energy density]Let $n := \dim M$. Let $\mathfrak{X}(M)$ denote the real [vector space](/page/Vector%20Space) of smooth vector fields on $M$. Let $\nabla^M$ and $\nabla^N$ denote the [Levi-Civita connections](/page/Levi-Civita%20Connection) of $(M,g)$ and $(N,h)$, respectively. Let $R^M$ and $R^N$ denote the [curvature tensors](/page/Riemann%20Curvature%20Tensor) of $(M,g)$ and $(N,h)$, with the convention \begin{align*} R(A,B)C := \nabla_A\nabla_BC - \nabla_B\nabla_AC - \nabla_{[A,B]}C. \end{align*} With this convention, define the Ricci tensor of $(M,g)$ by \begin{align*} \operatorname{Ric}_g(X,Y) := \sum_{j=1}^n g(R^M(X,e_j)e_j,Y), \end{align*} where $(e_1,\dots,e_n)$ is any local $g$-orthonormal frame. Let $\nabla^u$ denote the pullback connection on $u^*TN \to M \times [0,T)$, so that $\nabla^u_X$ is the spatial covariant derivative along $u$ for $X \in \mathfrak{X}(M)$ and $\nabla^u_{\partial_t}$ is the covariant time derivative along $u$. For each fixed $t \in [0,T)$, define \begin{align*} (\nabla du_t)(X,Y) := \nabla^u_X(du_t(Y)) - du_t(\nabla^M_XY) \end{align*} for smooth vector fields $X,Y \in \mathfrak{X}(M)$. The tension field is \begin{align*} \tau_g(u) = \sum_{i=1}^n (\nabla du_t)(e_i,e_i), \end{align*} where $(e_1,\dots,e_n)$ is any local $g$-orthonormal frame. Fix $(p,t) \in M \times [0,T)$. Choose a local $g$-orthonormal frame $(e_1,\dots,e_n)$ near $p$ such that $\nabla^M_{e_i}e_j(p)=0$ for all $i,j$. Define, at the fixed time $t$, the vector $u_i \in T_{u(p,t)}N$ by \begin{align*} u_i := du_t(e_i). \end{align*} Define the vector $u_{ij} \in T_{u(p,t)}N$ by \begin{align*} u_{ij} := (\nabla du_t)(e_i,e_j). \end{align*} Define the energy density function $e(u): M \times [0,T) \to \mathbb{R}$ by \begin{align*} e(u)(x,s) := \frac{1}{2}|du_s|_{g,h}^2, \end{align*} where $(x,s) \in M \times [0,T)$ and $|du_s|_{g,h}$ denotes the Hilbert-Schmidt norm of $du_s:T_xM \to T_{u(x,s)}N$ induced by $g_x$ and $h_{u(x,s)}$. In the chosen orthonormal frame at $(p,t)$, this means \begin{align*} e(u)(p,t)=\frac{1}{2}\sum_{i=1}^n |u_i|_h^2. \end{align*} At $p$, the normal-frame condition gives \begin{align*} \partial_t e(u) = \sum_{i=1}^n h(\nabla^u_{e_i}\tau_g(u),u_i) \end{align*} and \begin{align*} \Delta_g e(u) = \sum_{i,j=1}^n |u_{ij}|_h^2 +\sum_{i,j=1}^n h(\nabla^u_{e_j}\nabla^u_{e_j}u_i,u_i). \end{align*} Since $\tau_g(u)=\sum_{j=1}^n u_{jj}$ at $p$, subtracting gives \begin{align*} (\partial_t-\Delta_g)e(u) = -\sum_{i,j=1}^n |u_{ij}|_h^2 +\sum_{i,j=1}^n h\left(\nabla^u_{e_i}u_{jj}-\nabla^u_{e_j}\nabla^u_{e_j}u_i,u_i\right). \end{align*} We use the following pointwise commutation formula, proved from the curvature definitions at the normal frame. [claim:Commute the second covariant derivatives of $du_t$] At $p$, for each $i,j \in \{1,\dots,n\}$, \begin{align*} \nabla^u_{e_i}u_{jj}-\nabla^u_{e_j}\nabla^u_{e_j}u_i = R^N(u_i,u_j)u_j - du_t(R^M(e_i,e_j)e_j). \end{align*} [/claim] [proof] Because $\nabla^M_{e_i}e_j(p)=0$, the definitions of $u_i$ and $u_{jj}$ give \begin{align*} \nabla^u_{e_i}u_{jj}=\nabla^u_{e_i}\nabla^u_{e_j}u_j. \end{align*} Also, \begin{align*} \nabla^u_{e_j}\nabla^u_{e_j}u_i=\nabla^u_{e_j}\nabla^u_{e_j}du_t(e_i) \end{align*} at $p$. For the pullback connection, the curvature identity says \begin{align*} \nabla^u_{e_i}\nabla^u_{e_j}u_j-\nabla^u_{e_j}\nabla^u_{e_i}u_j-\nabla^u_{[e_i,e_j]}u_j = R^N(u_i,u_j)u_j. \end{align*} Since the Levi-Civita connection of $M$ is torsion-free, $[e_i,e_j]=\nabla^M_{e_i}e_j-\nabla^M_{e_j}e_i$, and the normal-frame condition makes $[e_i,e_j](p)=0$. It remains to account for the fact that $u_j=du_t(e_j)$ has a domain vector field in its argument. We derive the needed instance of the [Ricci Identity](/page/Ricci%20Identity) for the section $du_t$ of $T^*M\otimes u_t^*TN$. Let $Z \in \mathfrak{X}(M)$ be a smooth vector field. Since $du_t(Z)$ is a section of $u_t^*TN$, the curvature of the pullback connection gives \begin{align*} (\nabla^u_{e_i}\nabla^u_{e_j}-\nabla^u_{e_j}\nabla^u_{e_i}-\nabla^u_{[e_i,e_j]})(du_t(Z))=R^N(u_i,u_j)du_t(Z). \end{align*} Expanding the covariant derivatives of $du_t(Z)$ and using the definition of the curvature tensor on $TM$ gives \begin{align*} ((\nabla^u_{e_i}\nabla du_t)-(\nabla^u_{e_j}\nabla du_t))(e_j,Z) =R^N(u_i,u_j)du_t(Z)-du_t(R^M(e_i,e_j)Z) \end{align*} at $p$, where the negative sign is the curvature action on the covariant $T^*M$ slot. Taking $Z=e_j$ gives \begin{align*} (\nabla^u_{e_i}\nabla du_t)(e_j,e_j)-(\nabla^u_{e_j}\nabla du_t)(e_i,e_j) =R^N(u_i,u_j)u_j-du_t(R^M(e_i,e_j)e_j). \end{align*} Because the frame is normal at $p$, the covariant derivatives of the frame vectors vanish at $p$, so this identity is exactly \begin{align*} \nabla^u_{e_i}u_{jj}-\nabla^u_{e_j}\nabla^u_{e_j}u_i =R^N(u_i,u_j)u_j-du_t(R^M(e_i,e_j)e_j). \end{align*} This gives the displayed identity with the domain-curvature sign fixed by the convention for $R^M$. [/proof] Pairing the claim with $u_i$, summing over $i,j$, using the curvature symmetry \begin{align*} h(R^N(A,B)B,A)=-h(R^N(A,B)A,B), \end{align*} and using the definition $\operatorname{Ric}_g(X,Y)=\sum_{j=1}^n g(R^M(X,e_j)e_j,Y)$ gives \begin{align*} (\partial_t-\Delta_g)e(u) = -\sum_{i,j=1}^n |u_{ij}|_h^2 -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j) -\sum_{i,j=1}^n h\bigl(R^N(u_i,u_j)u_i,u_j\bigr), \end{align*} With the stated curvature convention, the sectional curvature of the plane spanned by linearly independent $X,Y \in T_qN$ is \begin{align*} K_N(X,Y)=\frac{h(R^N(X,Y)Y,X)}{|X|_h^2|Y|_h^2-h(X,Y)^2}. \end{align*} Thus nonpositive sectional curvature is equivalent, by the curvature symmetry $h(R^N(X,Y)Y,X)=-h(R^N(X,Y)X,Y)$, to \begin{align*} h(R^N(X,Y)X,Y) \geq 0 \end{align*} for all $X,Y \in T_qN$ and all $q \in N$.[/step]

[guided]We now compute the identity that drives the estimate. Let $n := \dim M$, and let $\mathfrak{X}(M)$ denote the real vector space of smooth vector fields on $M$. Let $R^M$ and $R^N$ denote the [curvature tensors](/page/Riemann%20Curvature%20Tensor) of $(M,g)$ and $(N,h)$, with the convention \begin{align*} R(A,B)C := \nabla_A\nabla_BC - \nabla_B\nabla_AC - \nabla_{[A,B]}C. \end{align*} With this same convention, the Ricci tensor of $(M,g)$ is \begin{align*} \operatorname{Ric}_g(X,Y) := \sum_{j=1}^n g(R^M(X,e_j)e_j,Y), \end{align*} where $(e_1,\dots,e_n)$ is any local $g$-orthonormal frame. The pullback connection $\nabla^u$ is taken on $u^*TN \to M \times [0,T)$, so it includes both spatial covariant derivatives and the covariant time derivative $\nabla^u_{\partial_t}$ along $u$. The covariant derivative of $du_t$ is defined by \begin{align*} (\nabla du_t)(X,Y) := \nabla^u_X(du_t(Y)) - du_t(\nabla^M_XY) \end{align*} for vector fields $X,Y \in \mathfrak{X}(M)$. This is a section of $T^*M \otimes T^*M \otimes u_t^*TN$. Its trace with respect to $g$ is the tension field \begin{align*} \tau_g(u) = \sum_{i=1}^n (\nabla du_t)(e_i,e_i), \end{align*} where $(e_1,\dots,e_n)$ is a local $g$-orthonormal frame. Fix a point $(p,t) \in M \times [0,T)$. We choose the frame so that $\nabla^M_{e_i}e_j(p)=0$ for every $i,j$. This choice is only a pointwise simplification; all quantities in the final identity are tensorial, so the identity obtained at $p$ is independent of the chosen frame. Define the vector $u_i \in T_{u(p,t)}N$ by \begin{align*} u_i := du_t(e_i). \end{align*} Define the vector $u_{ij} \in T_{u(p,t)}N$ by \begin{align*} u_{ij} := (\nabla du_t)(e_i,e_j). \end{align*} At the point $p$, the energy density is \begin{align*} e(u)=\frac{1}{2}\sum_{i=1}^n h(u_i,u_i). \end{align*} Differentiate in time. Since the connection on $N$ is torsion-free and $\partial_t u=\tau_g(u)$ by the harmonic map heat flow equation, \begin{align*} \partial_t e(u) = \sum_{i=1}^n h(\nabla^u_{\partial_t}u_i,u_i). \end{align*} Because the [Levi-Civita connection](/page/Levi-Civita%20Connection) on $N$ is torsion-free, the second fundamental covariant derivative of the map $u$ is symmetric in the two input directions. Here the input directions are the coordinate vector $\partial_t$ on $[0,T)$ and the spatial vector $e_i$ on $M$, and their product bracket vanishes, so \begin{align*} \nabla^u_{\partial_t}u_i = \nabla^u_{e_i}\partial_t u. \end{align*} Since $u$ solves [harmonic map heat flow](/page/Harmonic%20Map%20Heat%20Flow), $\partial_t u = \tau_g(u)$. Hence \begin{align*} \partial_t e(u) = \sum_{i=1}^n h(\nabla^u_{e_i}\tau_g(u),u_i). \end{align*} Next compute the Laplacian at $p$. Because the frame is normal at $p$, \begin{align*} \Delta_g e(u) = \sum_{j=1}^n e_j e_j\left(\frac{1}{2}\sum_{i=1}^n h(u_i,u_i)\right). \end{align*} Using compatibility of $\nabla^N$ with $h$ and the normal-frame condition $\nabla^M_{e_i}e_j(p)=0$, this becomes \begin{align*} \Delta_g e(u) = \sum_{i,j=1}^n |u_{ij}|_h^2 +\sum_{i,j=1}^n h(\nabla^u_{e_j}\nabla^u_{e_j}u_i,u_i). \end{align*} Subtracting the Laplacian from the time derivative gives \begin{align*} (\partial_t-\Delta_g)e(u) = -\sum_{i,j=1}^n |u_{ij}|_h^2 +\sum_{i=1}^n h(\nabla^u_{e_i}\tau_g(u),u_i) -\sum_{i,j=1}^n h(\nabla^u_{e_j}\nabla^u_{e_j}u_i,u_i). \end{align*} The remaining task is to commute covariant derivatives. At the normal frame, $\tau_g(u)=\sum_{j=1}^n u_{jj}$, so the expression to identify is \begin{align*} \sum_{i,j=1}^n h\left(\nabla^u_{e_i}u_{jj}-\nabla^u_{e_j}\nabla^u_{e_j}u_i,u_i\right). \end{align*} We now derive the commutation formula with signs visible. At $p$, because $\nabla^M_{e_i}e_j(p)=0$, we have \begin{align*} \nabla^u_{e_i}u_{jj}=\nabla^u_{e_i}\nabla^u_{e_j}u_j. \end{align*} Also, \begin{align*} \nabla^u_{e_j}\nabla^u_{e_j}u_i=\nabla^u_{e_j}\nabla^u_{e_j}du_t(e_i). \end{align*} The curvature identity for the pullback connection gives \begin{align*} \nabla^u_{e_i}\nabla^u_{e_j}u_j-\nabla^u_{e_j}\nabla^u_{e_i}u_j-\nabla^u_{[e_i,e_j]}u_j =R^N(u_i,u_j)u_j. \end{align*} The Levi-Civita connection on $M$ is torsion-free, so $[e_i,e_j]=\nabla^M_{e_i}e_j-\nabla^M_{e_j}e_i$, and the normal-frame condition gives $[e_i,e_j](p)=0$. We must still commute the covariant derivative acting on the $T^*M$ slot of $du_t$. We derive the needed instance of the [Ricci Identity](/page/Ricci%20Identity) rather than hiding the sign in a convention. Let $Z \in \mathfrak{X}(M)$ be a smooth vector field. Since $du_t(Z)$ is a section of $u_t^*TN$, the curvature of the pullback connection gives \begin{align*} (\nabla^u_{e_i}\nabla^u_{e_j}-\nabla^u_{e_j}\nabla^u_{e_i}-\nabla^u_{[e_i,e_j]})(du_t(Z))=R^N(u_i,u_j)du_t(Z). \end{align*} Now expand each derivative of $du_t(Z)$ using the definition of $\nabla du_t$. The curvature acting on the target-valued component contributes $R^N(u_i,u_j)du_t(Z)$, while the curvature acting on the covariant $T^*M$ input contributes with the opposite sign. Thus, at $p$, \begin{align*} ((\nabla^u_{e_i}\nabla du_t)-(\nabla^u_{e_j}\nabla du_t))(e_j,Z) =R^N(u_i,u_j)du_t(Z)-du_t(R^M(e_i,e_j)Z). \end{align*} Taking $Z=e_j$ gives \begin{align*} (\nabla^u_{e_i}\nabla du_t)(e_j,e_j)-(\nabla^u_{e_j}\nabla du_t)(e_i,e_j) =R^N(u_i,u_j)u_j-du_t(R^M(e_i,e_j)e_j). \end{align*} Because the frame is normal at $p$, this is precisely \begin{align*} \nabla^u_{e_i}u_{jj}-\nabla^u_{e_j}\nabla^u_{e_j}u_i =R^N(u_i,u_j)u_j-du_t(R^M(e_i,e_j)e_j). \end{align*} This is the point where sign discipline matters. Pairing the target term with $u_i$ gives $h(R^N(u_i,u_j)u_j,u_i)$, while the final Bochner identity is written with $h(R^N(u_i,u_j)u_i,u_j)$. These differ by a minus sign because the curvature tensor is skew-adjoint in its last two slots: \begin{align*} h(R^N(A,B)B,A)=-h(R^N(A,B)A,B). \end{align*} Pairing the commutation identity with $u_i$ and summing over $i,j$ gives \begin{align*} \sum_{i=1}^n h(\nabla^u_{e_i}\tau_g(u),u_i) -\sum_{i,j=1}^n h(\nabla^u_{e_j}\nabla^u_{e_j}u_i,u_i) = -\sum_{i,j=1}^n h\bigl(R^N(u_i,u_j)u_i,u_j\bigr) -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j). \end{align*} The last equality uses the convention $\operatorname{Ric}_g(X,Y)=\sum_{j=1}^n g(R^M(X,e_j)e_j,Y)$ with respect to the orthonormal frame $(e_1,\dots,e_n)$. Therefore \begin{align*} (\partial_t-\Delta_g)e(u) = -\sum_{i,j=1}^n |u_{ij}|_h^2 -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j) -\sum_{i,j=1}^n h\bigl(R^N(u_i,u_j)u_i,u_j\bigr). \end{align*} This is the parabolic Bochner identity for harmonic map heat flow. With the stated convention, the sectional curvature of the plane spanned by linearly independent tangent vectors $X,Y \in T_qN$ is \begin{align*} K_N(X,Y)=\frac{h(R^N(X,Y)Y,X)}{|X|_h^2|Y|_h^2-h(X,Y)^2}. \end{align*} Since $h(R^N(X,Y)Y,X)=-h(R^N(X,Y)X,Y)$, nonpositive sectional curvature of $N$ means \begin{align*} h(R^N(X,Y)X,Y) \geq 0 \end{align*} for all tangent vectors $X,Y \in T_qN$.[/guided]

[step:Use nonpositive target curvature to discard the target term]For every $q \in N$ and every $X,Y \in T_qN$, nonpositive sectional curvature gives \begin{align*} h(R^N(X,Y)X,Y) \geq 0. \end{align*} Applying this with $X=u_i$ and $Y=u_j$ yields \begin{align*} -\sum_{i,j=1}^n h\bigl(R^N(u_i,u_j)u_i,u_j\bigr) \leq 0. \end{align*} Also, \begin{align*} -\sum_{i,j=1}^n |u_{ij}|_h^2 \leq 0. \end{align*} Hence the Bochner identity implies \begin{align*} (\partial_t-\Delta_g)e(u) \leq -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j). \end{align*}[/step]

[guided]The Bochner identity has three terms on the right-hand side: \begin{align*} (\partial_t-\Delta_g)e(u) = -\sum_{i,j=1}^n |u_{ij}|_h^2 -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j) -\sum_{i,j=1}^n h\bigl(R^N(u_i,u_j)u_i,u_j\bigr). \end{align*} The first term is nonpositive because it is the negative of a sum of squared $h$-norms. For the target curvature term, nonpositive sectional curvature of $N$ means that for every $q \in N$ and all $X,Y \in T_qN$, \begin{align*} h(R^N(X,Y)X,Y) \geq 0. \end{align*} Applying this pointwise at $q=u(p,t)$ with $X=u_i$ and $Y=u_j$ gives \begin{align*} -\sum_{i,j=1}^n h\bigl(R^N(u_i,u_j)u_i,u_j\bigr) \leq 0. \end{align*} Therefore both the second-derivative term and the target-curvature term can be discarded from above, leaving \begin{align*} (\partial_t-\Delta_g)e(u) \leq -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j). \end{align*}[/guided]

[step:Bound the domain Ricci contribution by compactness of $M$]For each $p \in M$, let $\lambda_{\min}(p)$ denote the smallest eigenvalue of the Ricci tensor $\operatorname{Ric}_g(p)$ as a symmetric [bilinear form](/page/Bilinear%20Form) with respect to $g_p$. Since $M$ is closed, the [continuous function](/page/Continuous%20Function) $p \mapsto \lambda_{\min}(p)$ attains its minimum. Define \begin{align*} \kappa_M := \max\left\{0,-\min_{p \in M}\lambda_{\min}(p)\right\}. \end{align*} Then, for every $p \in M$ and every $X \in T_pM$, \begin{align*} -\operatorname{Ric}_g(X,X) \leq \kappa_M |X|_g^2. \end{align*} At the fixed point $p$, diagonalize $\operatorname{Ric}_g(p)$ in a $g_p$-[orthonormal basis](/page/Orthonormal%20Basis) $(e_1,\dots,e_n)$. Then \begin{align*} -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j) = -\sum_{i=1}^n \operatorname{Ric}_g(e_i,e_i)\,|u_i|_h^2. \end{align*} By the definition of $\kappa_M$, each term satisfies $-\operatorname{Ric}_g(e_i,e_i) \leq \kappa_M$. Therefore \begin{align*} -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j) \leq \kappa_M \sum_{i=1}^n |u_i|_h^2. \end{align*} Since $e(u)=\frac{1}{2}\sum_{i=1}^n |u_i|_h^2$, we obtain \begin{align*} -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j) \leq 2\kappa_M e(u). \end{align*} Therefore \begin{align*} (\partial_t-\Delta_g)e(u) \leq 2\kappa_M e(u). \end{align*} Setting $C_M := 2\kappa_M$ gives the asserted inequality. The constant depends only on the Ricci curvature of $(M,g)$, hence only on the geometry of $M$.[/step]

[guided]The remaining term involves only the Ricci curvature of the domain and the spatial differential of $u$. For each $p \in M$, let $\lambda_{\min}(p)$ be the smallest eigenvalue of the symmetric bilinear form $\operatorname{Ric}_g(p)$ with respect to $g_p$. Since $M$ is closed, it is compact without boundary, and the continuous function $p \mapsto \lambda_{\min}(p)$ attains a minimum. Define \begin{align*} \kappa_M := \max\left\{0,-\min_{p \in M}\lambda_{\min}(p)\right\}. \end{align*} Then for every $p \in M$ and every $X \in T_pM$, \begin{align*} -\operatorname{Ric}_g(X,X) \leq \kappa_M |X|_g^2. \end{align*} At the fixed point $p$, choose the orthonormal frame $(e_1,\dots,e_n)$ to diagonalize $\operatorname{Ric}_g(p)$. This is allowed because $\operatorname{Ric}_g(p)$ is symmetric. In this frame, \begin{align*} -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j) = -\sum_{i=1}^n \operatorname{Ric}_g(e_i,e_i)|u_i|_h^2. \end{align*} The definition of $\kappa_M$ gives $-\operatorname{Ric}_g(e_i,e_i) \leq \kappa_M$ for each $i$, so \begin{align*} -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j) \leq \kappa_M\sum_{i=1}^n |u_i|_h^2. \end{align*} Since the energy density is $e(u)=\frac{1}{2}\sum_{i=1}^n |u_i|_h^2$, this becomes \begin{align*} -\sum_{i,j=1}^n \operatorname{Ric}_g(e_i,e_j)\,h(u_i,u_j) \leq 2\kappa_M e(u). \end{align*} Combining this estimate with the previous guided step yields \begin{align*} (\partial_t-\Delta_g)e(u) \leq 2\kappa_M e(u). \end{align*} Thus $C_M := 2\kappa_M$ proves the asserted inequality, and $C_M$ depends only on the Ricci curvature of $(M,g)$.[/guided]

[step:Specialize the estimate when the domain Ricci curvature is nonnegative]If $\operatorname{Ric}_g \geq 0$ as a quadratic form on $TM$, then $\lambda_{\min}(p) \geq 0$ for every $p \in M$, so $\kappa_M = 0$. The preceding estimate becomes \begin{align*} (\partial_t-\Delta_g)e(u) \leq 0. \end{align*} Thus the theorem holds with $C_M=0$ when $\operatorname{Ric}_g \geq 0$.[/step]

[guided]If $\operatorname{Ric}_g \geq 0$ as a quadratic form, then every eigenvalue of $\operatorname{Ric}_g(p)$ is nonnegative for every $p \in M$. Hence $\lambda_{\min}(p) \geq 0$ for all $p$, and the constant defined above satisfies $\kappa_M=0$. Substituting this into the estimate already proved gives \begin{align*} (\partial_t-\Delta_g)e(u) \leq 0. \end{align*} This is exactly the theorem's final assertion, with $C_M=0$.[/guided]